• March 19th 2009, 06:40 PM
nosh
Im horrible at this :( sorry about it being bold .. it wont unbold
How do you evaluate the following limits

1) lim x----> 3 [ ( l x-3 l ) / ( x-3 ) ]

2) lim x --->4/8 [ ( 3x^2 + x - 4 ) / (3x + 4 ) ]

3) lim x--->1 [ (x^2 + x -2 ) / ( l x -1 l ) ]

4)lim x--->-1 [ ( 2x^2 +5x + 3 ) / (x+1) ]

5) lim x---->0 { [ x+8] ^ 1/3 -2} / x
• March 19th 2009, 07:03 PM
Reckoner
Quote:

1. $\lim_{x\to3}\frac{\lvert x-3\rvert}{x-3}$
Consider the left- and right-sided limits. For $x<3,$ $\lvert x-3\rvert=-(x-3)$ and for $x\geq3,$ $\lvert x-3\rvert=x-3.$ So,

$\lim_{x\to3^-}\frac{\lvert x-3\rvert}{x-3}$

$\lim_{x\to3^-}\frac{-(x-3)}{x-3}=-1$

and

$\lim_{x\to3^+}\frac{\lvert x-3\rvert}{x-3}$

$\lim_{x\to3^+}\frac{x-3}{x-3}=1.$

What does this tell you about the two-sided limit?

Quote:

2. $\lim_{x\to4/8}\frac{3x^2+x-4}{3x+4}$
This just requires direct substitution. Did you write the problem correctly?

Quote:

3. $\lim_{x\to1}\frac{x^2+x-2}{\lvert x-1\rvert}$
$\lim_{x\to1}\frac{x^2+x-2}{\lvert x-1\rvert}$

$=\lim_{x\to1}\frac{(x-1)(x+2)}{\lvert x-1\rvert}$

Again, consider the one-sided limits.

Quote:

4. $\lim_{x\to-1}\frac{2x^2+5x+3}{x+1}$
$\lim_{x\to-1}\frac{2x^2+5x+3}{x+1}$

$=\lim_{x\to-1}\frac{(2x+3)(x+1)}{x+1}$

$=\lim_{x\to-1}(2x+3)$

Quote:

5. $\lim_{x\to0}\frac{(x+8)^{1/3}-2}x$
$\lim_{x\to0}\frac{(x+8)^{1/3}-2}x=\frac00,$ so we can use L'Hôpital's rule.

$\lim_{x\to0}\frac{(x+8)^{1/3}-2}x=\lim_{x\to0}\frac13(x+8)^{-2/3}$

$=\lim_{x\to0}\frac1{3(x+8)^{2/3}}$

Finish it.
• March 19th 2009, 07:13 PM
nosh
Thank you so much .
• March 19th 2009, 07:18 PM
Krizalid
For the fifth one just put $u^3=x+8.$